Simplify the following expression and state the condition under which the simplification is valid. $z = \dfrac{t^2 - 64}{t + 8}$
Solution: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = t$ $ b = \sqrt{64} = 8$ So we can rewrite the expression as: $z = \dfrac{({t} + {8})({t} {-8})} {t + 8} $ We can divide the numerator and denominator by $(t + 8)$ on condition that $t \neq -8$ Therefore $z = t - 8; t \neq -8$